Why is the group $GL(n, \mathbb{R})$ of dimension $n^{2}$?

  • 2
    $\begingroup$ The group $M(n, \mathbb{R})$ of $n\times n$ matrices has dimension $n^2$. Why? It's easy to show that $M(n, \mathbb{R})$ a vector space. Any element $\in M(n, \mathbb{R})$ can be written in terms of a sum of $n^2$ matrices $B_{ij}$ where the entry $i,j)$ of $B_{ij} = 1$, and zero otherwise. You can show that the set $\{B_{ij}\}$ is a basis of $M(n, \mathbb{R})$, and hence $M(n, \mathbb{R})$ has dimension $n^2$. $GL(n,\mathbb{R})$ is a subset of $M(n, \mathbb{R})$ under the determinant map. It has the same dimension by Steve's answer below. $\endgroup$
    – user2468
    Mar 6, 2012 at 21:04

3 Answers 3


It is an open submanifold of the set of $n$ by $n$ matrices, which is a vector space of dimension $n^2$.

  • $\begingroup$ Is there are more elementary way to see this though? $\endgroup$
    – 198203
    Mar 6, 2012 at 20:47
  • 7
    $\begingroup$ @198203: Isn't that plenty elementary already? It's about the most elementary imaginable argument that can reach the relevant definition of "dimension" at all. $\endgroup$ Mar 6, 2012 at 20:54
  • 1
    $\begingroup$ How many basis elements does an $n$ by $n$ matrix have? Certainly it must be $n^2$... take the obvious basis: a single entry in each matrix to be $1$ and all others to be $0$. Do this $n \cdot n$ times, one for each position in the matrix. $\endgroup$
    – Tyler
    Mar 6, 2012 at 20:55
  • 2
    $\begingroup$ Basically, the determinant is a continuous function from the $n\times n$ matrices to $\mathbb R$, and the invertible ones are the ones with non-zero determinant, so $GL(n)=det^{-1}(\mathbb {R}\setminus 0)$. Since $\mathbb R\setminus 0$ is an open subset of $\mathbb R$, $GL(n)$ has to be an open subset of the set of all matrices. $\endgroup$ Mar 6, 2012 at 22:24
  • 2
    $\begingroup$ @198203 Do you know the definition of the dimension of a manifold? Once you have the definition, the fact that an open submanifold of an $n$-manifold has dimension $n$ is essentially immediate. What's non-trivial and requires work is the proof that the notion of the dimension of a manifold is well-defined. $\endgroup$ Mar 6, 2012 at 22:53

You should notice that the determinant is a continuous map $f:M(n,\mathbb{R})\equiv\mathbb{R}^{n^2}\to \mathbb{R}$, $f(X)=\det(X)$

Then note that

$$GL(n, \mathbb{R})=f^{-1}(\mathbb{R}\setminus\{0\})$$

How the pre-image by continuous map of open set is an open set then $GL(n, \mathbb{R})$ is an open set of $\mathbb{R}^{n^2}$ whose dimension is $n^2$.

(I am here using that an open set of $\mathbb{R}^{k}$ has $k$-dimension.)

  • $\begingroup$ How is, by the way, the dimension of an open set defined ? I am wondering because i am familiar with dimensions of vectorspaces etc. but not this. $\endgroup$
    – user42761
    Nov 19, 2012 at 6:43
  • 1
    $\begingroup$ If a manifold as a topological space is locally homeomorphic to $R^n$ then its dimension is $n$. We can also think about the Haussdorf Dimension. $\endgroup$
    – checkmath
    Nov 23, 2012 at 18:04
  • $\begingroup$ Why does the open set of $R^{n^2}$ has the dimension $R^{n^2}$? If n=2, $\{(x,y,0,0):x,y \in \Bbb R\} $ has basis $(1,0,0,0)$ and $(0,1,0,0)$ . $\endgroup$
    – math112358
    Jan 22, 2020 at 16:09

I'm sorry people misunderstood your question. You are clearly asking about dimension in terms of linear algebra and not in terms of manifolds.

The set $GL(n)$ is not a subspace. To see this, take any invertible matrix $A$, then $-A$ is also invertible. But $A-A=0$ is obviously not invertible, thus $GL(n)$ is not a vector subspace for any $n$ so it makes no sense to talk about it's dimension in terms of independent vectors.

However, GL(n) is what is known as a submanifold which means that although $GL(n)$ is not a vector space, you can "locally" (i.e. around some neighborhood around the set) view it a vector space structure. But this is something you shouldn't worry about in linear algebra.

  • 1
    $\begingroup$ I doubt the OP is still wondering this after ~10 years later lol $\endgroup$ Jan 19 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.